User blog:Allam948736/Ordinal hyperoperators and BEAF - analysis
The ordinal \(\varepsilon_0\) can be thought of as \(\omega \uparrow\uparrow \omega\), which naturally begs the question: how would the higher hyperoperators (and BEAF) work with ordinals? It turns out that there are two different ways to interpret ordinal hyperoperators beyond \(\varepsilon_0\). In what I call the "non-climbing interpretation", the definition of \(\omega \uparrow\uparrow (\omega + 1)\) varies but is usually either \(\varepsilon_1\) or \(\omega^{\varepsilon_0 + 1}\) (which is the same as \(\varepsilon_0 \times \omega\). Continuing with this logic we find that for \(\alpha\ \ge \omega^\omega\), \(\omega \uparrow\uparrow \alpha\ = \varepsilon_\alpha\), and that \(\omega \uparrow\uparrow\uparrow \omega\ = \zeta_0\). Similarly, \(\omega \uparrow\uparrow\uparrow\uparrow \omega\ = \eta_0\), and in general w^^^...^^^w w/ n arrows is equal to phi(n-1, 0). This means that \(\Gamma_0\) is equal to \(\omega \uparrow^{\omega \uparrow^{\omega \uparrow^...}}\) or \(\{\omega, \omega, 1, 2\}\). However, there is another way to define ordinal hyperoperators, where instead of interpreting the climb from \(\varepsilon_0\) to \(\varepsilon_1\) as repeated exponentiation of \(\omega\), we imagine the +1 climbing up the infinite tower. In this interpretation, which I refer to as the climbing interpretation, we don't hit \(\omega \uparrow\uparrow (\omega + 1)\) until \(\varepsilon_{\omega^\omega}\), and \(\zeta_0\) is merely \(\omega \uparrow\uparrow \omega^2\), and by the time we reach \(\omega \uparrow\uparrow\uparrow \omega\) or \(\Gamma_0\), we have already transcended the whole idea of ordinal up-arrows in the non-climbing interpretation entirely. In the climbing interpretation, \(\omega \uparrow^n \omega\) is equal to phi(n-2, 0, 0) for n > 2. What if we used a base other than \(\omega\), say \(\varepsilon_0\)? It turns out that for \(\alpha\ \ge \omega^2\), \(\varepsilon_0 \uparrow\uparrow \alpha\) is equal to \(\omega \uparrow\uparrow \alpha\) if we assume the climbing interpretation. Thus \(\varepsilon_0 \uparrow\uparrow \omega^2 = \omega \uparrow\uparrow \omega^2 = \zeta_0\), and \(\varepsilon_0 \uparrow\uparrow\uparrow 2 = \varphi(\varepsilon_0, 0)\), the same as \(\omega \uparrow\uparrow\uparrow 3\). In the climbing interpretation, \(\{\omega, \omega, 1, 2\}\) is the Ackermann ordinal, which is \(\{\omega, \omega, 1, 1, 2\}\) in the non-climbing interpretation. In general, phi(1, 0, 0, 0, ... 0, 0, 0, 0) w/ n 0s is equal to \(\{\omega, \omega, 1, 1, 1, ..., 1, 1, 1, 2\}\) w/ n-1 1s in the non-climbing interpretation, or n-2 1s in the climbing interpretation. The expression using the non-climbing interpretation consistently has one more argument than the expression using the climbing interpretation, meaning the two interpretations catch each other at the small Veblen ordinal \(\vartheta(\Omega^\omega)\), which is equal to \(\{\omega, \omega(1)2\}\). The large Veblen ordinal, \(\vartheta(\Omega^\Omega)\), is equal to \(\{\omega, \omega, 2(1)2\}\). I currently plan for the farthest reaches of my notation to reach the ordinal \(\{\omega, 3(1)3\}\) (how I'm going to get that far is another story), but I have no idea how that would be expressed using ordinal collapsing functions. Next, the Bachmann-Howard ordinal \(\vartheta(\varepsilon_{\Omega+1})\) can be imagined as \(\omega \uparrow\uparrow \omega\ \&\ \omega\), or \(\{\omega, \omega(\varepsilon_0)2\}\), or a tetrational array of omega. The slow-growing hierarchy catching ordinal under the most common interpretation (or \(\psi(\Omega_\omega)\)) can be imagined as w & w & w & w & ........., or the limit of the array of operator. EDIT: It turns out that there is another interpretation that has \(\omega \uparrow\uparrow (\omega + 1)\) as far out as \(\varphi(\omega, 0)\). In this interpretation, \(\Gamma_0\) is only \(\omega \uparrow\uparrow \omega^2\), and \(\omega \uparrow\uparrow\uparrow \omega\) is probably the large Veblen ordinal, but I'm not really sure. Also, there are actually three different variations of the non-climbing interpretation, which are all identical by \(\varepsilon_{\varepsilon_0}\), which is \(\omega \uparrow\uparrow\uparrow 3\) in all three. Category:Blog posts